by Corry Shores
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[The following is summary. Boldface (except for metavariables), underlining, and bracketed commentary are my own. I highly recommend Agler’s excellent book. It is one of the best introductions to logic I have come across.]
Summary of
David W. Agler
Symbolic Logic: Syntax, Semantics, and Proof
Ch.2: Language, Syntax, and Semantics
2.5 Advanced Translation
Brief Summary:
We translate the following formulations in the following ways:
1) “Neither P nor Q” = ¬P∧¬Q
2) “Not both P and Q” = ¬(P∧Q)
3) “P only if Q” = P→Q
4) “P even if Q” = P, or = P∧(Q∨¬Q)
5) “not-P unless Q” = ¬P∨Q
6) “P unless Q” = ¬(P↔Q), or = (P∨Q)∧¬(P∧Q)
Summary
2.5 Advanced Translation
We now consider how to translate less standard expressions into the truth-functional operators we have been discussing.
“Neither P nor Q” = ¬P∧¬Q
Expression with “neither P nor Q” are rendered ¬P∧¬Q, because they are true “if and only if both ‘P’ and ‘Q’ are false” (52). So
Neither John nor Liz plays guitar
Could be
¬J∧¬L
(52)
“Not both P and Q” = ¬(P∧Q)
Expressions formulated as “not both P and Q” are rendered ¬(P∧Q), because they are true “so long as ‘P’ and ‘Q’ are not jointly true” (53). This means that ¬(P∧Q) is true in three cases: (a) when just the first term is true, (b) when just the second term is true, and (c) when both terms are true. So
Frank did not kiss both Corinne and George
Could be
¬(C∧G)
(53)
“P only if Q” = P→Q
Expressions formulated “P only if Q” are not translated as Q→P. There are two ways to understand why. The first involves the distinction between necessary conditions and sufficient conditions.
‘P’ is a sufficient condition for ‘Q’ when the truth of ‘P’ guarantees the truth of ‘Q.’
By contrast, ‘P’ is a necessary condition for ‘Q’ when the falsity of ‘P’ guarantees the falsity of ‘Q.’
(52, underlined boldface mine)
So now we will see why in P→Q, “ ‘P’ is a sufficient condition for ‘Q,’ while ‘Q’ is a necessary condition for ‘P’ ” (53). Agler has us consider these sentences:
If Toronto is the largest city in Canada (‘P’), then Toronto is the largest city in Ontario (‘Q’).
Toronto is the largest city in Canada (‘P’).
Therefore, Toronto is largest city in Ontario (‘Q’).
(54)
[So consider the claim that “If Toronto is the largest city in Canada (‘P’), then Toronto is the largest city in Ontario (‘Q’).” If something is the largest member of a whole thing, then it is the largest member of any subdivision of that thing. So we might say that the truth of “Toronto is the largest city in Canada” guarantees the truth of “Toronto is the largest city in Ontario”. And recall also that ‘P’ is a sufficient condition for ‘Q’ when the truth of ‘P’ guarantees the truth of ‘Q.’ Thus in these example sentences,] “Notice that the truth of ‘P’ in the above argument guarantees the truth of ‘Q.’ Thus, ‘P’ is sufficient for ‘Q’ ” (54a). Now we consider instead the following sentences. [We notice that the first sentence is the same. But this time we negate the consequent and thereby negate the antecedent. But the insight remains the same, namely, that if something is the largest in the whole, then it is the largest in any of the parts. Thus if something is not the largest thing in one of the parts, then it is not the largest thing in the whole.]
If Toronto is the largest city in Canada (‘P’), then Toronto is the largest city in Ontario (‘Q’).
Toronto is not the largest city in Ontario (‘¬Q’).
Therefore, Toronto is not the largest city in [Canada] (‘¬P’).
(54, note, the bracketed insertion replaces “Ontario”)
[Recall before that ‘P’ is a necessary condition for ‘Q’ when the falsity of ‘P’ guarantees the falsity of ‘Q.’ But in this example, we need to think of the terms in the other order. Now, Q’s falsity guarantees P’s falsity. So in other words, The only way P can be true is if Q is true. In other words, only if Q (is true), then P (is true); or P only if Q. The prior situation was that P’s being true means that Q is true.]
Notice that the falsity of ‘Q’ in the above argument (as represented by the second premise) guarantees the falsity of ‘P’ (as represented by the conclusion). Thus, ‘Q’ is necessary for ‘P.’ Now consider that ‘P only if Q’ says that in order for ‘P’ to be true, ‘Q’ needs to be true. That is, ‘P only if Q’ says that ‘Q’ is a necessary condition for ‘P.’ Thus, ‘P only if Q’ should be translated as ‘P→Q.’
(54)
So that is the first way to explain why ‘P only if Q’ should be translated as ‘P→Q.’ The second way has us begin by noting the one case where ‘P only if Q’ is false, namely, when P is true and Q is false [for, here we have P is true, which we are saying can only happen if Q is true, which it is not in this instance.] So now we need a truth-functional operator where the whole proposition is true except when P is true and Q is false. This of course is the case for the material conditional operator. Agler illustrates. Consider one of his examples:
Ryan will let Daniel live only if Daniel wins the lottery.
(54)
This should be translated R→D because “P only if Q” should be translated P→Q. [Let “Ryan will let Daniel live” be R and “Daniel wins the lottery” be D. When would this be false? Let us look at the four possibilities. 1) R=t and D=t. Suppose Daniel wins the lottery and Ryan lets Daniel live. Is R→D false? No, because Daniel won the lottery, so the condition was met. 2) R=t and D=f. Now suppose that Daniel does not win the lottery and Ryan does still let Daniel live. Does that make R→D false? Yes, because we said Ryan would let Daniel live only if Daniel wins the lottery. But Daniel did not win it, so Ryan should not let Daniel live. 3) R=f and D=t. Suppose instead that Daniel does win the lottery, but Ryan does not let Daniel live. Does this make R→D false? We might think so. Here the condition was met, that Daniel wins the lottery. But what is this a condition for? It is a condition for Ryan letting Daniel live. It does not necessarily say anything about the situation where Ryan does not let Daniel live. In other words, it is possible that Ryan will not let Daniel live even if Daniel wins the lottery, because we stipulated that the only situations where Ryan does let Daniel live is when Daniel does win the lottery. Here Daniel did not win it, so we draw no conclusion about what Ryan does. 4) R=f and D=f. Does this make R→D false? No, for the same reason as above. When Daniel does not win, that places no restrictions on the situation where Ryan does not let Daniel live. So in other words, whatever comes after the “only if” should be the consequent and not the antecedent, because this best suits the truth table assignments.]
“P even if Q”
= P
or
= P∧(Q∨¬Q)
Consider the following statement:
Stock prices will go up even if people buy more stocks.
(55)
[We seem to be saying that stock prices are going to go up, period, and it does not matter whether or not people buy more stocks. This means when we translate this sentence symbolically, we could if we wanted leave out the second part of the sentence, since it has absolutely no bearing on the truth of the overall proposition. And recall we are dealing with truth-functionality, which means the truth of the whole is dependent on the truth of the parts. So if there is a part whose truth does not affect the truth of the whole, it is not truth-functionally related and thus can be excluded. Or instead, if we need still to communicate the original stated meaning for one reason or another, we would say that stocks will rise and either people will buy them or they will not.]
‘P even if Q’ is true if and only if ‘P’ is true. That is, ‘P even if Q’ says ‘P regardless of Q,’ and so the truth (or falsity) of ‘P even if Q’ entirely depends upon whether ‘P’ is true and is independent of whether ‘Q’ is true or false. Given that this is the case, there are two ways to translate ‘P even if Q.’ First, if the goal of a translation into a formal language is merely to capture the conditions under which a proposition is true, then we can disregard ‘Q’ altogether and translate ‘P even if Q’ as simply ‘P.’ However, if the goal of a translation is to preserve what is expressed, then we can translate ‘P even if Q’ as ‘P∧(Q∨¬Q)’.
(55)
“not-P unless Q” = ¬P∨Q.’
“P unless Q”
= (P∨Q)∧¬(P∧Q)
or
= ¬(P↔Q)
There are two ways to translate “P unless Q”, namely,
(P∨Q)∧¬(P∧Q) or ¬(P↔Q).
This is a very tricky formulation, so to understand better why we translate it in these ways, we should first consider a similar formulation that is actually “not-P unless Q”, for instance:
You will not win the lottery unless you acquire a ticket.
(55)
The only truth assignment where this would be false is if you did not buy the ticket but you still won the lottery.
[Let us say that “win the lottery” = P and thus “not win the lottery” is ¬P. And “acquire a ticket” is Q. There are four cases. 1) P=t (and thus ¬P=f; you do win the lottery, because it is false that you did not win) and Q=t. Suppose that you did buy the ticket and you did win the lottery. This means that the condition was met, and thus the whole statement is true. 2) P=t (and again ¬P=f for the same reason) and Q=f. Suppose that you did not get a ticket but you still won the lottery. The whole sentence is then false, because it stated that you need a ticket to win, but here you did not get one. 3) P=f (and thus ¬P=t; you did not win the lottery, because it is true that you did not win) and Q=t. Suppose that you acquire the ticket but you did not win. The whole sentence is still true, because we did not say that buying the ticket guarantees the win. It is just a precondition. Thus Q being true is not a sufficient condition for P being true. 4) P=f (and thus ¬P=t, for the same reason) and Q=f. Suppose lastly that you did not buy the ticket and you did not win. You did not meet the condition, so the sentence should still be true.]
If we let ‘P’ stand for You will win the lottery and ‘Q’ stand for You will acquire a lottery ticket, a translation of (11E) [You will not win the lottery unless you acquire a ticket] will be a proposition that is false just in the case that ‘P’ is true and ‘Q’ is false (and true in all others). Since ‘¬P∨Q’ is false only when ‘P’ is true and ‘Q’ is false, ‘¬P∨Q’ is a translation of (11E) [You will not win the lottery unless you acquire a ticket]. Thus, we can translate propositions like ‘not-P unless Q’ as ‘¬P∨Q.’
(56, bracketed insertions mine)
Agler then examines cases where “P unless Q” seems “to say something stronger (56). [Here also, we are not negating the P.] He has us consider this sentence:
John is at the party unless Liz called him.
(56)
This sentence says that “John is pretty much assured to be at the party, and the only thing that is going to stop him from being there is Liz’s call” (56). We will evaluate the truth value assignments for this, but we first skip the situation where they are both true. So call “John is at the party” P, and “Liz called John” Q. 1) P=t and Q=t ? [Skip for now]. 2) P=t and Q=f. Suppose John is at the party, but Liz did not call him. The sentence is true, because the only thing that would make John not go is if Liz called. But she did not call. 3) P=f and Q=t. This is also true, because here Liz did call him, and thus, as the sentence states, John is not at the party. 4) P=f and Q=f. Supposedly John will be at the party, except when Liz does call. But here John is not at the party, and Liz did not call. So under this truth-assignment, the sentence would be false.
But how do we evaluate the first truth-assignment where John is at the party and Liz did call him? Some say that the sentence would be true. For, “she may have called John to tell him to have a great time at the party” (57). But Agler finds this evaluation “to be unnatural and to depend upon ambiguous (and sometimes unarticulated) aspects of the sentence” (57). To illustrate, he has us consider this sentence:
I will take the job unless I get another offer.
(57)
Now suppose that both parts of the sentence are true. [Normally we would think that if he gets another offer, then he will not take the job.] And suppose further that we still want to say that the whole sentence is true. Then this sentence does not state what we really mean. In that case, we mean that we will take the job unless we get a better offer. Thus it is more natural to say that for “P unless Q”, it is false if both parts are true (57).
[So again, “we let ‘P’ stand for John is at the party and ‘Q’ stand for Liz called John. Given our truth assignments, we could make this table for the evaluation of the sentence:
We might note that the evaluation is the opposite of the biconditional. Thus] one way to represent this schema is with
¬(P↔Q)
, since it also is true and false for the exact same truth-assignments.
Also recall from section 2.4 that we had the same truth assignment for exclusive disjunction:
So we can also interpret it as P⊕Q. Or, we can interpret it as
(P∨Q)∧¬(P∧Q)
since this also has the same evaluation (57).
[I am a little confused by Agler’s next point. He says that we can translate ‘P unless Q’ as ‘P∨Q’. Perhaps this is because ‘not-P unless Q’, we said, was translatable as ‘¬P∨Q’ (see page 56). Then he writes, ‘not-P unless Q’ is best translated as ‘P∨Q,’ So I am confused. Let me quote:]
Knowing when to translate ‘P unless Q’ as ‘P∨Q’ as opposed to ‘¬(P↔Q)’ is a somewhat complicated affair. We suggest that as a rule of thumb, ‘P unless Q’ is translated as ‘¬(P↔Q)’ while ‘not-P unless Q’ is best translated as ‘P∨Q,’ but it is best to proceed in a step-by-step fashion like the method used above.
(57)
Agler, David. Symbolic Logic: Syntax, Semantics, and Proof. New York: Rowman & Littlefield, 2013.
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